POJ 3007 Organize Your Train part II map+string

map+string

 

/*Organize Your Train part IITime Limit: 1000MS Memory Limit: 65536K Total Submissions: 4370 Accepted: 1224 DescriptionRJ Freight, a Japanese railroad company for freight operations has recently constructed exchange lines at Hazawa, Yokohama. The layout of the lines is shown in Figure 1.Figure 1: Layout of the exchange linesA freight train consists of 2 to 72 freight cars. There are 26 types of freight cars, which are denoted by 26 lowercase letters from "a" to "z". The cars of the same type are indistinguishable from each other, and each car's direction doesn't matter either. Thus, a string of lowercase letters of length 2 to 72 is sufficient to completely express the configuration of a train.Upon arrival at the exchange lines, a train is divided into two sub-trains at an arbitrary position (prior to entering the storage lines). Each of the sub-trains may have its direction reversed (using the reversal line). Finally, the two sub-trains are connected in either order to form the final configuration. Note that the reversal operation is optional for each of the sub-trains.For example, if the arrival configuration is "abcd", the train is split into two sub-trains of either 3:1, 2:2 or 1:3 cars. For each of the splitting, possible final configurations are as follows ("+" indicates final concatenation position):[3:1] abc+d cba+d d+abc d+cba [2:2] ab+cd ab+dc ba+cd ba+dc cd+ab cd+ba dc+ab dc+ba [1:3] a+bcd a+dcb bcd+a dcb+aExcluding duplicates, 12 distinct configurations are possible.Given an arrival configuration, answer the number of distinct configurations which can be constructed using the exchange lines described above.InputThe entire input looks like the following.the number of datasets = m1st dataset 2nd dataset ... m-th datasetEach dataset represents an arriving train, and is a string of 2 to 72 lowercase letters in an input line.OutputFor each dataset, output the number of possible train configurations in a line. No other characters should appear in the output.Sample Input4aaabbaabcdabcdeSample Output161218SourceJapan 2006 Domestic*/#pragma warning(disable : 4786)#include <stdio.h>#include <string.h>#include <map>#include <string>#include <algorithm>using namespace std;map <string,int> mapExist;int main(void){int iLoop;int iNum;int iResult;int iStrLen;char acBuff[73];char acBuff1[73];char acBuffReverse[73];char acBuffReverse1[73];string strDataset; string strPart1;string strPart2;string strReversePart1;string strReversePart2;scanf("%d",&iNum);while (iNum--){iResult = 1;memset(acBuff1,0,73*sizeof(char));mapExist.clear();scanf("%s",acBuff);iStrLen = strlen(acBuff);strDataset = acBuff;mapExist[strDataset] = 1;for (iLoop = 0; iLoop < iStrLen; iLoop++ ){acBuffReverse[iLoop] = acBuff[iStrLen - 1-iLoop];}acBuffReverse[iStrLen] = 0;memcpy(acBuffReverse1,acBuffReverse,73);for (iLoop = 1; iLoop < iStrLen; iLoop++){acBuff1[iLoop-1] = acBuff[iLoop-1];strPart1 = acBuff1;strPart2 = &acBuff[iLoop];acBuffReverse1[iStrLen-iLoop] = 0;strReversePart1 = &acBuffReverse[iStrLen-iLoop];strReversePart2 = acBuffReverse1;strDataset = strPart2+strPart1;if (0 == mapExist[strDataset]){iResult++;mapExist[strDataset] = 1;}if (strPart1 != strReversePart1){strDataset = strReversePart1+strPart2;if (0 == mapExist[strDataset]){iResult++;mapExist[strDataset] = 1;}strDataset = strPart2+strReversePart1;if (0 == mapExist[strDataset]){iResult++;mapExist[strDataset] = 1;}}if (strPart2 != strReversePart2){strDataset = strPart1+strReversePart2;if (0 == mapExist[strDataset]){iResult++;mapExist[strDataset] = 1;}strDataset = strReversePart2+ strPart1;if (0 == mapExist[strDataset]){iResult++;mapExist[strDataset] = 1;}}if (strPart1 != strReversePart1&&strPart2 != strReversePart2){strDataset = strReversePart1+strReversePart2;if (0 == mapExist[strDataset]){iResult++;mapExist[strDataset] = 1;}strDataset = strReversePart2+strReversePart1;if (0 == mapExist[strDataset]){iResult++;mapExist[strDataset] = 1;}}}printf("%d/n",iResult);} return 0;}